Work for processes A and B are -909 kJ and -232.3. And heat for processes A and B are 859 and 0. Respectively.
It is the branch of science in which heat and other forms of energy are studied.
Given
Two processes A and B between the same end states, 1 and 2,
[tex]\begin{aligned} \rm &P_{1} = 1\ bar\ ,\ &P_{2} = 10\ bar\\\rm &V_{1} = 1\ m^{3}\ ,\ &V_{2} = 0.1\ m^{3} \\\rm &U_{1} = 400\ kJ\ ,\ &U_{2} = 450\ kJ \\\end{aligned}[/tex]
Process A: Constant-volume process from state 1 to a pressure of 10 bar, followed by a constant pressure process to state 2.
Process B: Process from 1 to 2 during which the pressure-volume relation is PV = constant.
All the calculation is done for the unit mass.
a. P-V graph is shown in the diagram.
b. Work for process A will be
[tex]\rm W_{A} = contant\ volume\ work\ +\ constant\ pressure\ work\\W_{A} = P_{1} dV + P_{2} dV\\W_{A} = 1*101 (V_{2} - V_{1} ) + 10*101* (V_{2} - V_{1} )\\W_{A} = 100 (1-1) + 1000 (0.1-1)\\W_{A} = -900[/tex]
The negative (-) sign represent work is done on the system.
Work for process B will be
[tex]\rm W_{B} = Work\ done\ for\ isothermal\ process\\ W_{B} = P_{1} V_{1} Ln \dfrac{V_{2} }{V_{1} } \\ W_{B} = 1*101 * 1 *Ln\dfrac{0.1}{1} \\ W_{B} = -232.3[/tex]
c. Heat = change in internal energy + work done.
Change in internal energy = 450 - 400 = 50kJ
Heat for process A will be
[tex]\rm \Delta H_{A} = \Delta U\ +\ W_{A} \\\Delta H_{A} = 50\ - 909\\\Delta H_{A} = -859[/tex]
The negative (-) sign represents the heat is given to the system.
Heat for process B will be
[tex]\rm H_{B} = W_{B} \\H_{B} = -232.3[/tex],
because the change in internal energy is zero for the isothermal process.
Thus, work for processes A and B are -909 kJ and -232.3 kJ. And heat for processes A and B are 859 and 0. Respectively.
More about the thermodynamic link is given below.
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